Proving that $f(x)=x^2$ is uniformly continuous on any closed interval, but not on $\mathbb{R}$

Question

Let $f: [a,b]\rightarrow\mathbb{R}$, $x\mapsto x^2$ where $a,b\in\mathbb{R}$

Without loss of generality, assume $|a|>|b|$

Let $\epsilon>0, x$ and $y\in[a,b]$ and $\delta=\min\{1,\frac{\epsilon}{2|a|+1}\}$

Further suppose $|x-y|<\delta$

Then $|x+y|=|x+x-x+y|\le |x+x|+|y-x|=|2x|+|x-y|<2|a|+\delta$

So $|f(x)-f(y)|=|x^2-y^2|=|x-y||x+y|<\delta(2|a|+\delta)\le(2|a|+1)\delta\le\frac{(2|a|+1)\epsilon}{2|a|+1}=\epsilon$

$\therefore\space f$ is continuous on $[a.b]$

Is my approach correct? I can see why this would not work on $\mathbb{R}$ (would not be able to bound $|2x|$ for example) but I am not able to visualize it. Any tips on how to see it graphically at least?

Answer 1

What you did is correct.

In order to prove that $f$ is not uniformly continuous, use the fact that, given $\delta>0$,$$\lim_{x\to+\infty}\bigl(f(x+\delta)-f(x)\bigr)=+\infty.$$